A) 1
B) 2
C) 3
D) None of these
Correct Answer: D
Solution :
\[h(x)=[f{{(x)}^{2}}+g{{(x)}^{2}}]\] \[\Rightarrow \] \[h\,(x)=2f(x)f(x)+2g(x)g(x)\] \[\Rightarrow \] \[h\,(x)=2f(x)g(x)+2g(x)f\,(x)\] \[\left[ \begin{align} & \because g(x)=f\,(x) \\ & \Rightarrow g(x)=f\,(x) \\ \end{align} \right]\] \[\therefore \] \[h\,(x)=2f(x)g(x)+2g(x)(-f(x))\] \[[\because f\,(x)=-f(x)]\] \[\Rightarrow \]\[h\,(x)=0\] \[\Rightarrow \] \[h(x)=C,a\] a constant for all \[x\in R\] \[\Rightarrow \] \[h(x)=Cx+{{C}_{1}}\] \[\Rightarrow \] \[h(0)={{C}_{1}}\]and \[h(1)=C+{{C}_{1}}\] \[\Rightarrow \] \[2={{C}_{1}}\]and \[8=C+{{C}_{1}}\] \[\Rightarrow \] \[{{C}_{1}}=2\]and \[C=6\] \[\therefore \] \[h(x)=6x+2\] \[\Rightarrow \] \[h(2)=6\times 2+2=14\]
You need to login to perform this action.
You will be redirected in
3 sec
